[alis88]

Hello all

Please can someone explain the following pinacol coupling mechanism to form the diol in the reaction:

http://img132.imageshack.us/my.php?image=pinacolco8.jpg

The reagents are SmI2, t-BuOH and THF.

Is THF the solvent?

Does t-BuOH provide the hydrogens?

And SmI2 creates the radical???

Thank you in advance.

Regards

Alis

[dfodor]

Yes, SmI2 is a single-electron reducing agent, THF is the solvent.
A proton donor is required for the reaction but t-BuOH is strange for me. Are you sure there is nothing else?

[alis88]

Yes, I am sure there is nothing else.

The reagents are SmI2, t-BuOH, and THF at -78 C to rt.

The reaction can be found in Tetrahedron Letters, article "Synthesis of N-heterocyclic diols by diastereoselective pinacol coupling reactions" by Sandeep Handa, Manpreet S. Kachala and Sarah R. Lowe.

Please could you help me with the mechanism?

Kind regards

Alis

[Mitsunobo]

SmI2 attacks the carbonyl oxygen creating a radical which can be delocalized to the carbonyl carbon. Sm can coordinate the 2 oxygen the same way Mg does in the pinacol reaction.

The carbonyl carbon radicals joins to form a bond.

The hydroxyls will be formed after Sm is displaced by H+ which will be taken from tBuOH.

Hope that helps.